Questions

A calorimeter of water equivalent 20 g contains 180g of water at ${25}^{0}$C. ‘m’ grams of steam at ${100}^{0}$C is mixed in it till the temperature of the mixture is ${31}^{0}$C. The value of ‘m’ is close to (Latent heat of water = 540 cal ${g}^{-1}$, specific heat of water = 1 cal ${g}^{-1}$ ${}^{0}{C}^{-1}$)

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By Expert Faculty of Sri Chaitanya

a

2

b

4

c

3.2

d

2.6

NEW

detailed solution

Correct option is A

Heat gained by cold body = heat lost by hot body⇒20+180×1×31-25=m(540)+m×1×100-31⇒1200=m(609)⇒m≈2gms

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40^{o}C. When m gram of ice at -10^{o}C is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^{o}C. It is known that specific heat capacity of the liquid changes with temperature as $\mathrm{S}=\left(1+\frac{\mathrm{\theta}}{500}\right)\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$where $\mathrm{\theta}$ is temperature in ^{o}C. The specific heat capacity of ice, water and the calorimeter remains constant and values are ${\mathrm{S}}_{\text{ice}}=0.5\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$; ${\mathrm{S}}_{\text{water}}=1.0\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$ and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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